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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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At $300\: K$, the standard enthalpies of formation of $C_6H_5COOH(s), CO_2(g)\: and\: H_2O(l)\: are\: -408, -393\: and\: -286 \: kJ/mol$ respectively. Calculate the heat of combustion of benzoic acid at constant volume and at constant pressure. $(R=8.31\: J/Kmol)$

$\begin {array} {1 1} (A)\;-3202.25 \: kJ/mol & \quad (B)\;3202.25 \: kJ/mol \\ (C)\;-3209.25\: kJ/mol & \quad (D)\;3209.25\: kJ/mol \end {array}$

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$C_6H_5COOH + \large\frac{13}{2}$$ O_2 \rightarrow 7CO_2 + 3H_2O$
$q(p) = \Delta H = 7(-393) + 3(-286) – (-408) – 0$
$ = -2571 – 858 + 408 = -3201\: kJ/mol$
$ \Delta n = 7-\large\frac{13}{2}$$ = 0.5$
$q_{ (V)} = \Delta E = \Delta H - \Delta nRT = -3201 – 0.5 \times 8.31 \times 10-3 \times 300 = -3202.25\: kJ/mol$
Ans : (A)
answered Mar 19, 2014 by thanvigandhi_1
 

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